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Rearrange 5y-3x-4=0 to get y=mx+c? Help please watch

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    (Original post by Naruke)
    this is gcse maybe even KS3
    That's what you think
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    (Original post by Pinkberry_y)
    That's what you think
    lmao
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    (Original post by vector12)
    Thank you. I wasn't sure what happened to the 0, but it seems that once you move the -4 to the other side and it becomes +4 to cancel it out, that then makes the 0 disappear from the equation and gets rid of it?

    So basically, the 0 is gotten rid of once you move one of the subjects to the right side of the equation, and it is just a placeholder after the = sign?
    You're welcome!
    Yeah you can think of it in a way of 'zeroing' a unit on a side so when you add 4

    5y - 3x - 4 = 0
    (+4) 5y - 3x -4 + 4 = 0 + 4 (+4)
    obviously 4-4 = 0. So it essentially 'zeroes' it on the y side - you do this till you get the y on it's own.
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    (Original post by Pinkberry_y)
    That's what you think
    It is.
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    (Original post by vector12)
    How do you do it? I can do it normally, just I cannot rearrange it with the 0.

    Can someone also explain to me what the 0 does please, and if it means anything?

    Thanks!
    Add 3x to the right hand side and left hand side. Add 4 to the right hand side and left hand side.

    You should end up with 5y=3x+4. because the 3x's and 4's subtract and give zero on the left hand side and the 3x and 4 add to the zero on the right hand side.

    Divide everything by 5.

    You should end up with y=3/5x + 4/5.

    Makes sense?
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    where does it say that?

    (Original post by RDKGames)
    Whether you agree with me or not is irrelevant. Posting full solutions on this forum is against its rules. Just hint the OP in the right direction next time.
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    (Original post by zainyyyyy)
    where does it say that?
    Sticky at the top of the forum. The one you're supposed to read before posting.

    http://www.thestudentroom.co.uk/show...9#post64637319
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    (Original post by zainyyyyy)
    where does it say that?
    http://www.thestudentroom.co.uk/show...9#post64637319

    I'm guessing you didn't read the big red "Please Read - Posting Guide" thing at the top of the page...?
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    (Original post by Pinkberry_y)
    Y = 3/5x +20
    BTW it's against the rules to just give the answer. Also, I would guess that it's super against the rules to give the wrong answer.
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    (Original post by anosmianAcrimony)
    BTW it's against the rules to just give the answer. Also, I would guess that it's super against the rules to give the wrong answer.
    hahahhahahahhahaha
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    (Original post by anosmianAcrimony)
    BTW it's against the rules to just give the answer. Also, I would guess that it's super against the rules to give the wrong answer.
    (Original post by Naruke)
    hahahhahahahhahaha
    Can someone tell me how my answer is wrong seeing as this is a learning process for all.
    FYI I meant 3x/5
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    (Original post by Pinkberry_y)
    Can someone tell me how my answer is wrong seeing as this is a learning process for all.
    FYI I meant 3x/5
    So you want me to teach you?
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    (Original post by Pinkberry_y)
    Can someone tell me how my answer is wrong seeing as this is a learning process for all.
    FYI I meant 3x/5
    3x/5 is correct, but 4/5 is not 20.
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    Well y = mx+c = mx + c+ 0, but 0=5y-3x-4, so, substituting, y = mx+c+5y-3x-4.
    The rest is easy.
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    (Original post by EricPiphany)
    Well y = mx+c = mx + c+ 0, but 0=5y-3x-4, so, substituting, y = mx+c+5y-3x-4.
    The rest is easy.
    Er, no.

    That would get him y=\frac{1}{4}(3-m)x+\frac{1}{4}(4-c) which is not what we want.
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    (Original post by RDKGames)
    3x/5 is correct, but 4/5 is not 20.
    Ohhhhh OMG I multiplied by accident.
    (Original post by Naruke)
    So you want me to teach you?
    See I did know what I was doing
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    (Original post by RDKGames)
    Er, no.

    That would get him y=\frac{1}{4}(3-m)x+\frac{1}{4}(4-c) which is not what we want.
    But you forget that 5y-3x-4 = 0.
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    (Original post by Pinkberry_y)
    Ohhhhh OMG I multiplied by accident.


    See I did know what I was doing
    if you say so bae
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    (Original post by RDKGames)
    Er, no.

    That would get him y=\frac{1}{4}(3-m)x+\frac{1}{4}(4-c) which is not what we want.
    It was a joke.
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    (Original post by Zacken)
    It was a joke.
    Of course
    Spoiler:
    Show
    ...but what if it secretly wasn't? :shock:
 
 
 
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